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Electronic Configuration and Oxidation States

What will hap...

Question

What will happen if the solution of potassium chromate reacts with excess amount of nitric acid?

C r

reduces in the oxidation state +3 from

C r {O_{4}}^{2 -}

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C r

oxidizes in the oxidation state +7 from

C r {O_{4}}^{2 -}

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{C r}^{+ 3}

and

C r_{2} {O_{7}}^{2 -}

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C r_{2} {O_{7}}^{2 -}

and

H_{2} O

will be formed

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Solution

The correct option is D $C r_{2} {O_{7}}^{2 -}$ and $H_{2} O$ will be formed
When potassum chromate reacts with dilute nitric acid ${C r}_{2} O_{7}^{2 -}$ and $H_{2} O$ are formed.
$2 K_{2} C r O_{4} + 2 H N O_{3} \to K_{2} C r_{2} O_{7} + 2 K N O_{3} + H_{2} O$

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{M n O}_{4}^{-} {, C r O}_{4}^{2 -}, {C r}_{2} O_{7}^{2 -}

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S O_{4}^{2 -}

C r_{2} O_{7}^{2 -}

{C r}_{2} O_{7}^{2 -} H^{+} - - \to C r

{C r}_{2} O_{7}^{2 -}

{C O}_{2}

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(H_{2} {C O}_{3} ⇌ H^{+} + {H C O}_{3}^{-})

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{C r}_{2} O_{7}^{2 -}

{C r}_{2} O_{7}^{2 -}

C r

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